Mettl Equation Quiz 3

I. x² – 17.5x + 69 = 0
x² – 11.5x – 6x + 69 = 0
x(x – 11.5) – 6(x – 11.5) = 0
(x – 11.5)(x – 6) = 0
x = 11.5, 6

II. 2y² – 25y + 78 = 0
2y² – 13y – 12y + 78 = 0
y (2y – 13) – 6(2y – 13) = 0
(2y – 13)(y – 6) = 0
y = 6.5, 6
For x = 11.5 and y = 6.5, 6 x > y
For x = 6, and y = 6.5, 6 x ≤ y

Therefore, the relationship can’t be established

I. 2x² + 11√3x + 45 = 0
2x² + 6√3x + 5√3x + 45 = 0
2x(x + 3√3) + 5√3 (x + 3√3) = 0
(2x + 5√3)(x + 3√3) = 0
x = – 3√3, \frac{-5\sqrt{3}}{2}

II. y² + 8√3y + 45 = 0
y² + 5√3y + 3√3y + 45 = 0
y(y + 5√3) + 3√3 (y + 5√3) = 0
(y + 5√3)(y + 3√3) = 0
y = – 5√3, – 3√3

For x = -3√3 and y = -3√3 x = y
For x = -3√3 and y = -5√3 x > y
For x = \frac{-5 \sqrt3}{2} and y = -5√3, – 3√3 x > y
Therefore, x ≥ y

I. x² + 9x + 20 = 0
x² + 5x + 4x + 20 = 0
x(x + 5) + 4(x + 5) = 0
(x + 4)(x + 5) = 0
x = -4, -5

II. y² + y – 12 = 0
y2 + 4y - 3y - 12 = 0
y (y + 4) - 3(y + 4) = 0
(y + 4)(y - 3) = 0
y = -4, 3

For x = -4 and y = – 4, x = y
For x = -4, y = 3 , x < y
For x = -5, y = -4 or 3 , x < y

Therefore, x ≤ y

I: 3x² – 243 = 0
3x² = 243, x² = 81 , x = ± 9
II: 12y⁵ + 110y⁴ = 0
12y5 = –110y4,
y = \frac{-110}{12} = -9.16 , 0

For x = 9, and y = -9.16 , 0 x > y
For x = -9, and y = -9.16, x > y
For x = -9, and y = 0 x < y

Therefore, the relationship cannot be established

I. 15x² - 11x + 2 = 0
15x² - 5x - 6x + 2 = 0
5x (3x - 1) - 2 (3x - 1) = 0
(3x - 1) (5x - 2) = 0
x =\frac{1}{3}, \frac{2}{5}

II. 10y² - 9y + 2 = 0
10y² - 5y - 4y + 2 = 0
5y (2y - 1) -2 (2y - 1) = 0
(2y - 1) (5y - 2) = 0
y = \frac{1}{2}, \frac{2}{5}
Therefore, x ≤ y

I. 10x² – 7x +1 = 0
10x² – 5x – 2x +1 = 0
5x (2x – 1) – 1 (2x – 1) = 0
(5x – 1) (2x – 1) = 0
x = \frac{1}{5}, \frac{1}{2}

II. 35y² – 12y + 1 = 0
35y² – 7y – 5y + 1 = 0
7y (5y – 1) – 1(5y – 1) = 0
(7y – 1) (5y – 1) = 0
y = \frac{1}{7}, \frac{1}{5}

Therefore, x ≥ y

I. x² – 6x = 7
x² – 6x – 7 = 0
x² – 7x + x – 7=0
x (x – 7) + 1 (x – 7) =0
(x + 1) (x – 7) =0
x = – 1, 7

II. 2y² + 13y + 15 = 0
2y² + 10y + 3y +15 = 0
2y (y + 5) +3 (y + 5) =0
(2y + 3) (y + 5) = 0
y = -5, \frac{-3}{2}

Hence, x > y

I. x² + 9x + 18 = 0
x² + 6x + 3x + 18 = 0
x (x + 6) + 3 (x + 6) =0
(x + 6) (x + 3) =0
x = - 6, -3

II. y² - 13y + 40 = 0
y² - 8y - 5y + 40 = 0
y(y - 8) - 5 (y - 8) =0
(y - 8) (y - 5) = 0
y = 8, 5

Hence, x < y

I. 14x² - 5 √15x - 90 = 0
14x² - (12 √15 - 7√15) x - 90 = 0
14x² - 12√15x + 7 √15x - 90 = 0
2x (7x - 6 √15) + √15 (7x - 6 √15) = 0
(2x + √15) (7x - 6√15) = 0
x = \frac{-\sqrt{15}}{2}, \frac{6\sqrt{15}}{7}

II. 6y² + √21y – 21 = 0
6y² + (3√21 – 2√21) y – 21 = 0
6y² + 3 √21y – 2 √21y – 21 = 0
3y (2y + √21) – √21 (2y + √21) = 0
(3y – √21) (2y + √21) = 0
y = \frac{\sqrt{21}}{3}, \frac{- \sqrt{21}}{2}

Hence, no relation can be established.

I. 4x² - 52x + 133 = 0
4x² - 14x - 38x + 133 = 0
2x(2x - 7) - 19(2x - 7) = 0
(2x - 7)(2x - 19) = 0
x = 9.5, 3.5

II. 4y² - 24y + 35 = 0
4y² - 14y - 10y + 35= 0
2y(2y - 7) - 5(2y - 7) = 0
(2y - 5)(2y - 7) = 0
y = 3.5, 2.5

For x = 3.5, and y = 3.5 x = y
For x = 3.5, or 9.5 and y = 2.5
x > y

Therefore, x ≥ y